91Ó°ÊÓ

It is the rule to check the aromaticity of the compound by knowing the number of delocalized electrons and the n value must be a whole number like when 6 pi electrons are present n=1.

$\begin{array}{c}\mathbf{4}\mathbf{n}\mathbf{+}\mathbf{2}\mathbf{=}\mathbf{6}\\ \mathbf{4}\mathbf{n}\mathbf{=}\mathbf{6}\mathbf{-}\mathbf{2}\\ \mathbf{4}\mathbf{n}\mathbf{=}\mathbf{4}\\ \mathbf{n}\mathbf{=}\mathbf{1}\end{array}$

It’s a heterocyclic five-membered aromatic compound the N and S are the heteroatoms present in the ring with two lone pairs on S and one lone pair on N and it is basically having two double bonds which show delocalization with a lone pair of nitrogen and one lone pair of sulfur.

Ï€-electron distribution is shown:

Thiazolium ring has 2 double bonds that are 4 pi electrons and the nitrogen has 1 lone pair in which1 electron takes part in delocalization. Sulfur have one electron which takes part in delocalization so 6 π-electron are present that is n=1 a whole number that 4n+2 rule is followed so thiazolium is an aromatic ring.